Find $\lim_{x\to 5}\dfrac{x+2}{x^2-5}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{3}{25}$ (Choice B) B $\dfrac{7}{20}$ (Choice C) C $-\dfrac{7}{25}$ (Choice D) D The limit doesn't exist
Solution: Let's try to find the limit using direct substitution. $\begin{aligned} \lim_{x\to 5}\dfrac{x+2}{x^2-5}&=\dfrac{5+2}{(5)^2-5} \\\\ &=\dfrac{7}{25-5} \\\\ &=\dfrac{7}{20} \end{aligned}$ We got a finite number. Since $\dfrac{x+2}{x^2-5}$ is continuous across its domain, we can determine that $\lim_{x\to 5}\dfrac{x+2}{x^2-5}$ is indeed equal to $\dfrac{7}{20}$. In conclusion, $\lim_{x\to 5}\dfrac{x+2}{x^2-5}=\dfrac{7}{20}$.